Let $A(a,\frac{a}{\sqrt{3}}),B(0,0),C(c,0)$. Point $H(a,\sqrt{3}(c-a))$. We use $w=\sqrt{3(c-a)^{2}+a^{2}}$. $\triangle ABH$, bissector $y-\frac{a}{\sqrt{3}}=\sqrt{3}(x-a)$ cuts the bissector $\sqrt{3}(c-a)x-ay=w(x-a)$ in the point $O_{1}(\frac{2a^{2}+\sqrt{3}aw}{6a+\sqrt{3}w-3c},\frac{a(6c+\sqrt{3}w-6a)}{3(2\sqrt{3}a-\sqrt{3}c+w)})$. $\triangle BCH$, bissector $y=-\frac{x-c}{\sqrt{3}}$ cuts the bissector $\sqrt{3}(c-a)x-ay=wy$ in the point $O_{2}(\frac{c(a+w)}{3c+w-2a},\frac{\sqrt{3}c(c-a)}{3c+w-2a})$. The lines $CO_{1}$ and $AO_{2}$ have slopes $m_{CO_{1}}$ and $m_{AO_{2}}$. The angle $\alpha$ of this lines: $\tan \alpha=\frac{m_{CO_{1}}-m_{AO_{2}}}{1+m_{CO_{1}} \cdot m_{AO_{2}}}=1$, so $\alpha=45^{\circ}$.